CODEWITHSUNDEEP
carrayspointersstruct
user840834
user840834

Reputation: 21

incompatible types between pointer to struct and its array data member

I am a C beginner, and I am getting an error when compiling the following:

typedef struct myStructType {
    uint8_t myArray[6];
    uint8_t x;
}myStruct;

myStruct *pmyStruct;

(*pmyStruct).myArray={1,2,3,4,5,6};   //getting error here
(*pmystruct).x=3;

The error is as follows: incompatible types when assigning to type ‘uint8_t[6]’ from type ‘uint8_t *’

Upvotes: 2

Views: 1927

Answers (5)

AudioDroid
AudioDroid

Reputation: 2322

If you want to keep things simple at first, and you don't want to deal with malloc and free yet, you can also first create a struct-object, then fill it with data, and then set a pointer to it:

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

typedef struct myStruct {
    uint8_t myArray[6];
    uint8_t x;
} myStruct_t;

int main()
{
   //create obj
   myStruct_t myObj;

   //fill it with data
   myObj.myArray[0] = 1;
   myObj.myArray[1] = 2;
   myObj.myArray[2] = 3;
   myObj.myArray[3] = 4;
   myObj.myArray[4] = 5;
   myObj.myArray[5] = 6;

   myObj.x = 3;

   //set a pointer to it
   myStruct_t *pObj = &myObj;

   return 0;
}

Upvotes: 0

jonsca
jonsca

Reputation: 10381

An array can be initialized when it is declared, so you could have:

int myarr[] = {1,2,3,4,5,6};

but you can't do this with it being a member of your struct.

For your purposes here just initialize it using a loop

 int i;
 for (i = 1;i<7;++i)
    pmyStruct->myArray = i;

(noting that (*pmyStruct).myArray can use the arrow notation pmyStruct->myArray)

For more general sets of numbers memcpy() can be used instead of the for loop

(as two of the other posters noted, you must allocate space for your struct before doing any of this, and deallocate that space when you are finished)

Upvotes: 0

pmg
pmg

Reputation: 108968

As you are using C99 (the uint8_t and // comment say so), you can try compound literals

#include <inttypes.h>
#include <stdlib.h>

struct myStructType {
  uint8_t myArray[6];
  uint8_t x;
};

int main(void) {
  struct myStructType *pmyStruct;
  pmyStruct = malloc(sizeof *pmyStruct);
  if (pmyStruct) {
    *pmyStruct = (struct myStructType){{1, 2, 3, 4, 5, 6}, 3}; // <== THIS
    free(pmyStruct);
  }
  return 0;
}

Upvotes: 2

BlackJack
BlackJack

Reputation: 2876

There are quite a few issues here.

First of all, make sure you have #include <stdint.h> in your file because some platforms will have an issue with uint8_t.

Secondly, when you create a struct pointer, you must allocate space for that struct. This can be done with the following code:

myStruct *pmyStruct = malloc(sizeof(struct myStructType));

As others have mentioned, you can only use {1,2,3...} when initializing the array. So to add elements to the array, you need:

for (int i = 0; i < 6; i++)
    pmyStruct->myArray[i] = i+1;

When you have a struct pointer, you access it's elements with the -> operand. So, your last line would be:

pmyStruct->x=3;

and finally, you need a main method. Putting it all together:

#include <stdlib.h>
#include <stdint.h>

typedef struct myStructType {
    uint8_t myArray[6];
    uint8_t x;
}myStruct;

int main()
{
    myStruct *pmyStruct = malloc(sizeof (struct myStructType));

    for (int i = 0; i < 6; i++)
        pmyStruct->myArray[i] = i+1;

    pmyStruct->x=3;

return 0;
}

Once you're done with everything, be sure to free the allocated space with free(pmyStruct); And pay attention to capitalizations! pmystruct is NOT the same as pmyStruct.

Best of luck.

Upvotes: 2

zw324
zw324

Reputation: 27180

(*pmyStruct).myArray={1,2,3,4,5,6};

You can only initialize an array this way. Use it like this is after it is declared will result in an error.

Your question is similar to this post.

Upvotes: 1

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